Integrand size = 15, antiderivative size = 80 \[ \int \frac {\sqrt {2+b x}}{x^{11/2}} \, dx=-\frac {(2+b x)^{3/2}}{9 x^{9/2}}+\frac {b (2+b x)^{3/2}}{21 x^{7/2}}-\frac {2 b^2 (2+b x)^{3/2}}{105 x^{5/2}}+\frac {2 b^3 (2+b x)^{3/2}}{315 x^{3/2}} \] Output:
-1/9*(b*x+2)^(3/2)/x^(9/2)+1/21*b*(b*x+2)^(3/2)/x^(7/2)-2/105*b^2*(b*x+2)^ (3/2)/x^(5/2)+2/315*b^3*(b*x+2)^(3/2)/x^(3/2)
Time = 0.06 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.60 \[ \int \frac {\sqrt {2+b x}}{x^{11/2}} \, dx=\frac {\sqrt {2+b x} \left (-70-5 b x+3 b^2 x^2-2 b^3 x^3+2 b^4 x^4\right )}{315 x^{9/2}} \] Input:
Integrate[Sqrt[2 + b*x]/x^(11/2),x]
Output:
(Sqrt[2 + b*x]*(-70 - 5*b*x + 3*b^2*x^2 - 2*b^3*x^3 + 2*b^4*x^4))/(315*x^( 9/2))
Time = 0.15 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.08, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {55, 55, 55, 48}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {\sqrt {b x+2}}{x^{11/2}} \, dx\) |
\(\Big \downarrow \) 55 |
\(\displaystyle -\frac {1}{3} b \int \frac {\sqrt {b x+2}}{x^{9/2}}dx-\frac {(b x+2)^{3/2}}{9 x^{9/2}}\) |
\(\Big \downarrow \) 55 |
\(\displaystyle -\frac {1}{3} b \left (-\frac {2}{7} b \int \frac {\sqrt {b x+2}}{x^{7/2}}dx-\frac {(b x+2)^{3/2}}{7 x^{7/2}}\right )-\frac {(b x+2)^{3/2}}{9 x^{9/2}}\) |
\(\Big \downarrow \) 55 |
\(\displaystyle -\frac {1}{3} b \left (-\frac {2}{7} b \left (-\frac {1}{5} b \int \frac {\sqrt {b x+2}}{x^{5/2}}dx-\frac {(b x+2)^{3/2}}{5 x^{5/2}}\right )-\frac {(b x+2)^{3/2}}{7 x^{7/2}}\right )-\frac {(b x+2)^{3/2}}{9 x^{9/2}}\) |
\(\Big \downarrow \) 48 |
\(\displaystyle -\frac {(b x+2)^{3/2}}{9 x^{9/2}}-\frac {1}{3} b \left (-\frac {(b x+2)^{3/2}}{7 x^{7/2}}-\frac {2}{7} b \left (\frac {b (b x+2)^{3/2}}{15 x^{3/2}}-\frac {(b x+2)^{3/2}}{5 x^{5/2}}\right )\right )\) |
Input:
Int[Sqrt[2 + b*x]/x^(11/2),x]
Output:
-1/9*(2 + b*x)^(3/2)/x^(9/2) - (b*(-1/7*(2 + b*x)^(3/2)/x^(7/2) - (2*b*(-1 /5*(2 + b*x)^(3/2)/x^(5/2) + (b*(2 + b*x)^(3/2))/(15*x^(3/2))))/7))/3
Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp [(a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] /; FreeQ[{ a, b, c, d, m, n}, x] && EqQ[m + n + 2, 0] && NeQ[m, -1]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] - Simp[d*(S implify[m + n + 2]/((b*c - a*d)*(m + 1))) Int[(a + b*x)^Simplify[m + 1]*( c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && ILtQ[Simplify[m + n + 2], 0] && NeQ[m, -1] && !(LtQ[m, -1] && LtQ[n, -1] && (EqQ[a, 0] || (NeQ[ c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && (SumSimplerQ[m, 1] || !SumSimp lerQ[n, 1])
Time = 0.07 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.44
method | result | size |
gosper | \(\frac {\left (b x +2\right )^{\frac {3}{2}} \left (2 b^{3} x^{3}-6 b^{2} x^{2}+15 b x -35\right )}{315 x^{\frac {9}{2}}}\) | \(35\) |
orering | \(\frac {\left (b x +2\right )^{\frac {3}{2}} \left (2 b^{3} x^{3}-6 b^{2} x^{2}+15 b x -35\right )}{315 x^{\frac {9}{2}}}\) | \(35\) |
meijerg | \(-\frac {2 \sqrt {2}\, \left (-\frac {1}{35} b^{4} x^{4}+\frac {1}{35} b^{3} x^{3}-\frac {3}{70} b^{2} x^{2}+\frac {1}{14} b x +1\right ) \sqrt {\frac {b x}{2}+1}}{9 x^{\frac {9}{2}}}\) | \(47\) |
risch | \(\frac {2 b^{5} x^{5}+2 b^{4} x^{4}-b^{3} x^{3}+b^{2} x^{2}-80 b x -140}{315 x^{\frac {9}{2}} \sqrt {b x +2}}\) | \(50\) |
default | \(-\frac {2 \sqrt {b x +2}}{9 x^{\frac {9}{2}}}+\frac {b \left (-\frac {\sqrt {b x +2}}{7 x^{\frac {7}{2}}}-\frac {3 b \left (-\frac {\sqrt {b x +2}}{5 x^{\frac {5}{2}}}-\frac {2 b \left (-\frac {\sqrt {b x +2}}{3 x^{\frac {3}{2}}}+\frac {b \sqrt {b x +2}}{3 \sqrt {x}}\right )}{5}\right )}{7}\right )}{9}\) | \(75\) |
Input:
int((b*x+2)^(1/2)/x^(11/2),x,method=_RETURNVERBOSE)
Output:
1/315*(b*x+2)^(3/2)*(2*b^3*x^3-6*b^2*x^2+15*b*x-35)/x^(9/2)
Time = 0.09 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.52 \[ \int \frac {\sqrt {2+b x}}{x^{11/2}} \, dx=\frac {{\left (2 \, b^{4} x^{4} - 2 \, b^{3} x^{3} + 3 \, b^{2} x^{2} - 5 \, b x - 70\right )} \sqrt {b x + 2}}{315 \, x^{\frac {9}{2}}} \] Input:
integrate((b*x+2)^(1/2)/x^(11/2),x, algorithm="fricas")
Output:
1/315*(2*b^4*x^4 - 2*b^3*x^3 + 3*b^2*x^2 - 5*b*x - 70)*sqrt(b*x + 2)/x^(9/ 2)
Leaf count of result is larger than twice the leaf count of optimal. 428 vs. \(2 (71) = 142\).
Time = 15.22 (sec) , antiderivative size = 428, normalized size of antiderivative = 5.35 \[ \int \frac {\sqrt {2+b x}}{x^{11/2}} \, dx=\frac {2 b^{\frac {33}{2}} x^{7} \sqrt {1 + \frac {2}{b x}}}{315 b^{12} x^{7} + 1890 b^{11} x^{6} + 3780 b^{10} x^{5} + 2520 b^{9} x^{4}} + \frac {10 b^{\frac {31}{2}} x^{6} \sqrt {1 + \frac {2}{b x}}}{315 b^{12} x^{7} + 1890 b^{11} x^{6} + 3780 b^{10} x^{5} + 2520 b^{9} x^{4}} + \frac {15 b^{\frac {29}{2}} x^{5} \sqrt {1 + \frac {2}{b x}}}{315 b^{12} x^{7} + 1890 b^{11} x^{6} + 3780 b^{10} x^{5} + 2520 b^{9} x^{4}} + \frac {5 b^{\frac {27}{2}} x^{4} \sqrt {1 + \frac {2}{b x}}}{315 b^{12} x^{7} + 1890 b^{11} x^{6} + 3780 b^{10} x^{5} + 2520 b^{9} x^{4}} - \frac {80 b^{\frac {25}{2}} x^{3} \sqrt {1 + \frac {2}{b x}}}{315 b^{12} x^{7} + 1890 b^{11} x^{6} + 3780 b^{10} x^{5} + 2520 b^{9} x^{4}} - \frac {456 b^{\frac {23}{2}} x^{2} \sqrt {1 + \frac {2}{b x}}}{315 b^{12} x^{7} + 1890 b^{11} x^{6} + 3780 b^{10} x^{5} + 2520 b^{9} x^{4}} - \frac {880 b^{\frac {21}{2}} x \sqrt {1 + \frac {2}{b x}}}{315 b^{12} x^{7} + 1890 b^{11} x^{6} + 3780 b^{10} x^{5} + 2520 b^{9} x^{4}} - \frac {560 b^{\frac {19}{2}} \sqrt {1 + \frac {2}{b x}}}{315 b^{12} x^{7} + 1890 b^{11} x^{6} + 3780 b^{10} x^{5} + 2520 b^{9} x^{4}} \] Input:
integrate((b*x+2)**(1/2)/x**(11/2),x)
Output:
2*b**(33/2)*x**7*sqrt(1 + 2/(b*x))/(315*b**12*x**7 + 1890*b**11*x**6 + 378 0*b**10*x**5 + 2520*b**9*x**4) + 10*b**(31/2)*x**6*sqrt(1 + 2/(b*x))/(315* b**12*x**7 + 1890*b**11*x**6 + 3780*b**10*x**5 + 2520*b**9*x**4) + 15*b**( 29/2)*x**5*sqrt(1 + 2/(b*x))/(315*b**12*x**7 + 1890*b**11*x**6 + 3780*b**1 0*x**5 + 2520*b**9*x**4) + 5*b**(27/2)*x**4*sqrt(1 + 2/(b*x))/(315*b**12*x **7 + 1890*b**11*x**6 + 3780*b**10*x**5 + 2520*b**9*x**4) - 80*b**(25/2)*x **3*sqrt(1 + 2/(b*x))/(315*b**12*x**7 + 1890*b**11*x**6 + 3780*b**10*x**5 + 2520*b**9*x**4) - 456*b**(23/2)*x**2*sqrt(1 + 2/(b*x))/(315*b**12*x**7 + 1890*b**11*x**6 + 3780*b**10*x**5 + 2520*b**9*x**4) - 880*b**(21/2)*x*sqr t(1 + 2/(b*x))/(315*b**12*x**7 + 1890*b**11*x**6 + 3780*b**10*x**5 + 2520* b**9*x**4) - 560*b**(19/2)*sqrt(1 + 2/(b*x))/(315*b**12*x**7 + 1890*b**11* x**6 + 3780*b**10*x**5 + 2520*b**9*x**4)
Time = 0.05 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.70 \[ \int \frac {\sqrt {2+b x}}{x^{11/2}} \, dx=\frac {{\left (b x + 2\right )}^{\frac {3}{2}} b^{3}}{24 \, x^{\frac {3}{2}}} - \frac {3 \, {\left (b x + 2\right )}^{\frac {5}{2}} b^{2}}{40 \, x^{\frac {5}{2}}} + \frac {3 \, {\left (b x + 2\right )}^{\frac {7}{2}} b}{56 \, x^{\frac {7}{2}}} - \frac {{\left (b x + 2\right )}^{\frac {9}{2}}}{72 \, x^{\frac {9}{2}}} \] Input:
integrate((b*x+2)^(1/2)/x^(11/2),x, algorithm="maxima")
Output:
1/24*(b*x + 2)^(3/2)*b^3/x^(3/2) - 3/40*(b*x + 2)^(5/2)*b^2/x^(5/2) + 3/56 *(b*x + 2)^(7/2)*b/x^(7/2) - 1/72*(b*x + 2)^(9/2)/x^(9/2)
Time = 0.14 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.85 \[ \int \frac {\sqrt {2+b x}}{x^{11/2}} \, dx=-\frac {{\left (105 \, b^{9} - {\left (63 \, b^{9} + 2 \, {\left ({\left (b x + 2\right )} b^{9} - 9 \, b^{9}\right )} {\left (b x + 2\right )}\right )} {\left (b x + 2\right )}\right )} {\left (b x + 2\right )}^{\frac {3}{2}} b}{315 \, {\left ({\left (b x + 2\right )} b - 2 \, b\right )}^{\frac {9}{2}} {\left | b \right |}} \] Input:
integrate((b*x+2)^(1/2)/x^(11/2),x, algorithm="giac")
Output:
-1/315*(105*b^9 - (63*b^9 + 2*((b*x + 2)*b^9 - 9*b^9)*(b*x + 2))*(b*x + 2) )*(b*x + 2)^(3/2)*b/(((b*x + 2)*b - 2*b)^(9/2)*abs(b))
Time = 0.16 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.52 \[ \int \frac {\sqrt {2+b x}}{x^{11/2}} \, dx=-\frac {\sqrt {b\,x+2}\,\left (-\frac {2\,b^4\,x^4}{315}+\frac {2\,b^3\,x^3}{315}-\frac {b^2\,x^2}{105}+\frac {b\,x}{63}+\frac {2}{9}\right )}{x^{9/2}} \] Input:
int((b*x + 2)^(1/2)/x^(11/2),x)
Output:
-((b*x + 2)^(1/2)*((b*x)/63 - (b^2*x^2)/105 + (2*b^3*x^3)/315 - (2*b^4*x^4 )/315 + 2/9))/x^(9/2)
Time = 0.16 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.08 \[ \int \frac {\sqrt {2+b x}}{x^{11/2}} \, dx=\frac {2 \sqrt {x}\, \sqrt {b x +2}\, b^{4} x^{4}-2 \sqrt {x}\, \sqrt {b x +2}\, b^{3} x^{3}+3 \sqrt {x}\, \sqrt {b x +2}\, b^{2} x^{2}-5 \sqrt {x}\, \sqrt {b x +2}\, b x -70 \sqrt {x}\, \sqrt {b x +2}-2 \sqrt {b}\, b^{4} x^{5}}{315 x^{5}} \] Input:
int((b*x+2)^(1/2)/x^(11/2),x)
Output:
(2*sqrt(x)*sqrt(b*x + 2)*b**4*x**4 - 2*sqrt(x)*sqrt(b*x + 2)*b**3*x**3 + 3 *sqrt(x)*sqrt(b*x + 2)*b**2*x**2 - 5*sqrt(x)*sqrt(b*x + 2)*b*x - 70*sqrt(x )*sqrt(b*x + 2) - 2*sqrt(b)*b**4*x**5)/(315*x**5)